Find discrete antiderivative, i.e. $ f ( x ) = g ( x + 1 ) - g ( x ) $ for given $ f ( x ) $. 
Let $ f : \mathbb R \to \mathbb C $ be a given smooth function. Is there an easy way to find smooth function $ g ( x ) $ such that $ f ( x ) = g ( x + 1 ) - g ( x ) $?

I think it is not hard if $ f ( x ) $ is polynomial (we only have to find such function for $ f ( x ) = x ^ k $), but I can't find general solution for arbitrary function $ f $.
 A: Let $ h : [ 0 , 1 ) \to \mathbb C $ be any smooth function with
$$ \lim _ { x \to 1 ^ - } h ^ { ( n ) } ( x ) = h ^ { ( n ) } ( 0 ) + f ^ { ( n ) } ( 0 ) \tag {*} \label 0 $$
for all $ n \in \mathbb Z _ { 0 + } $. Define $ g : \mathbb R \to \mathbb R $ with
$$ g ( x ) =
\begin {cases}
h ( x - \lfloor x \rfloor ) + \sum \limits _ { n = 1 } ^ { \lfloor x \rfloor } f ( x - n ) & x \ge 0 \\
h ( x - \lfloor x \rfloor ) - \sum \limits _ { n = \lfloor x \rfloor + 1 } ^ 0 f ( x - n ) & x < 0
\end {cases} \tag {**} \label 1 $$
for all $ x \in \mathbb R $. Then it's straightforward to verify that $ g ( x + 1 ) - g ( x ) = f ( x ) $ for all $ x \in \mathbb R $. Smoothness of $ g $ is a consequence of \eqref{0}. There are uncountably many smooth functions $ h : [ 0 , 1 ) \to \mathbb C $ satisfying \eqref{0}, and thus you can find uncountably many suitable candidates for $ g $. Note that conversely, for any suitable $ g $, defining $ h = g | _ { [ 0 , 1 ) } $, we can see that $ h $ is smooth and satisfies \eqref{0} for all $ n \in \mathbb Z _ { 0 + } $. Then, using the fact that $ g ( x - \lfloor x \rfloor ) = h ( x - \lfloor x \rfloor ) $ and $ g ( x + 1 ) - g ( x ) = f ( x ) $ for all $ x \in \mathbb R $, we can show that \eqref{1} holds (by induction on $ \lfloor x \rfloor $ when $ x \ge 0 $ and on $ - \lfloor x \rfloor $ when $ x < 0 $). This shows that all the possible solutions are of the form presented above.
